Let B=[1, 3, α; 1, 2, 3; α, α, 4], α > 2 be the adjoint of a matrix A and |A|=2. Then [α, -2 α, α] B[α; -2 α; α] is equal to :

Solution

Maths sir · 4 steps

Find α from |adj A| = |A|², then pull α out of both vectors before you multiply.

1. Determinant of adjoint. For a 3 × 3 matrix, |adj A| = |A|². Here B = adj A and |A| = 2, so |B| = 4.

   |B| = 1(8-3α) - 3(4-3α) + α(α-2α) = -α² + 6α - 4 = 4

2. Find alpha. This gives α² - 6α + 8 = 0, that is (α-2)(α-4) = 0. Since α > 2, we get α = 4.
3. Factor and multiply. The column vector is α (1, -2, 1)ᵀ, and the row vector is its transpose, so the product is α² vᵀ B v with v = (1, -2, 1)ᵀ. With α = 4, Bv = (1-6+4, 1-4+3, 4-8+4)ᵀ = (-1, 0, 0)ᵀ, so vᵀ B v = -1.
4. So. The value is 16 × (-1) = -16, which is option (b).

The idea to keep. For an n × n matrix, |adj A| = |A|ⁿ⁻¹. When a vector is a multiple of a simple vector, pull the scalar out first. The arithmetic becomes much lighter.

Answer from the NTA final answer key. Solution by PrepWiser.